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🟥 Binary Tree Maximum Path Sum (#124)

🔗 Problem Link​

📋 Problem Statement​

A path in a binary tree is a sequence of nodes where each pair of adjacent nodes in the sequence has an edge connecting them. A node can only appear in the sequence at most once. Note that the path does not need to pass through the root.

The path sum of a path is the sum of the node's values in the path.

Given the root of a binary tree, return the maximum path sum of any non-empty path.

💡 Examples​

Example 1​

Input: root = [1,2,3]
Output: 6
Explanation: The optimal path is 2 -> 1 -> 3 with a path sum of 2 + 1 + 3 = 6.

Example 2​

Input: root = [-10,9,20,null,null,15,7]
Output: 42
Explanation: The optimal path is 15 -> 20 -> 7 with a path sum of 15 + 20 + 7 = 42.

🔑 Key Insights & Approach​

Core Observation: At each node, max path can:

  1. Go through node as split point (left + node + right)
  2. Continue from node to parent (node + max(left, right))

Track global max for case 1, return case 2 to parent.

Why Post-order DFS?

  • O(n) time, O(h) space
  • Need child results before processing node
  • Handle negative paths by using max(0, child)

Approaches:

  1. Post-order DFS with global max: O(n) time, O(h) space - optimal

Pattern: "Tree Path Sum" with split point consideration.

🐍 Solution: Python​

Approach: Post-order DFS with Global Max​

Time Complexity: O(n) | Space Complexity: O(h)

from typing import Optional

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

class Solution:
    def maxPathSum(self, root: Optional[TreeNode]) -> int:
        self.max_sum = float('-inf')

        def dfs(node):
            if not node:
                return 0

            # Get max path sum from left and right (ignore negative)
            left_max = max(dfs(node.left), 0)
            right_max = max(dfs(node.right), 0)

            # Update global max (path through current node)
            current_max = node.val + left_max + right_max
            self.max_sum = max(self.max_sum, current_max)

            # Return max path sum continuing to parent
            return node.val + max(left_max, right_max)

        dfs(root)
        return self.max_sum

🔵 Solution: Golang​

Approach: Post-order DFS​

Time Complexity: O(n) | Space Complexity: O(h)

import "math"

type TreeNode struct {
    Val   int
    Left  *TreeNode
    Right *TreeNode
}

func maxPathSum(root *TreeNode) int {
    maxSum := math.MinInt32

    var dfs func(node *TreeNode) int
    dfs = func(node *TreeNode) int {
        if node == nil {
            return 0
        }

        // Get max from left and right (ignore negative)
        leftMax := max(dfs(node.Left), 0)
        rightMax := max(dfs(node.Right), 0)

        // Update global max
        currentMax := node.Val + leftMax + rightMax
        if currentMax > maxSum {
            maxSum = currentMax
        }

        // Return max path continuing to parent
        return node.Val + max(leftMax, rightMax)
    }

    dfs(root)
    return maxSum
}

func max(a, b int) int {
    if a > b {
        return a
    }
    return b
}

📚 References​